Problem: You have found the following ages (in years) of all 4 seals at your local zoo: $ 32,\enspace 2,\enspace 4,\enspace 12$ What is the average age of the seals at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{32 + 2 + 4 + 12}{{4}} = {12.5\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $32$ years $19.5$ years $380.25$ years $^2$ $2$ years $-10.5$ years $110.25$ years $^2$ $4$ years $-8.5$ years $72.25$ years $^2$ $12$ years $-0.5$ years $0.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{380.25} + {110.25} + {72.25} + {0.25}} {{4}} $ $ {\sigma^2} = \dfrac{{563}}{{4}} = {140.75\text{ years}^2} $ The average seal at the zoo is 12.5 years old. The population variance is 140.75 years $^2$.